Q 11-04-012JEE MainEAMCET 2014 (Engineering)Medium
A body of mass $8$ kg is moved by a force $F = (3x)$ N, where $x$ is the distance covered. Its initial position is $x = 2$ m and its final position is $x = 10$ m. If the body is initially at rest, its final speed is
Answer: (C) $6\ \text{m/s}$
Write the acceleration as $v\dfrac{dv}{dx}$:
$$3x = 8v\frac{dv}{dx} \;\Rightarrow\; \int_2^{10} 3x\,dx = \int_0^v 8v\,dv$$
$$\frac{3}{2}(100 - 4) = 4v^2 \;\Rightarrow\; 144 = 4v^2 \;\Rightarrow\; v = 6\ \text{m/s}$$
(Equivalently, the work done $\int F\,dx = 144$ J equals the kinetic energy gained.)
Solution by Sreeraj P, M.Sc Physics